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Ex 11.4, 3 - 9y2 - 4x2 = 36 Find vertices, eccentricity - Hyperbola

Ex 11.4,  3 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.4,  3 - Chapter 11 Class 11 Conic Sections - Part 3


Transcript

Ex 11.4, 3 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y2 – 4x2 = 36 The given equation is 9y2 – 4x2 = 36 Divide whole equation by 36 ﷐9﷐𝑦﷮2﷯ − 4﷐𝑥﷮2﷯﷮36﷯ = ﷐36﷮36﷯ ﷐9﷐𝑦﷮2﷯﷮36﷯ − ﷐4﷮36﷯x2 = 1 ﷐﷐𝑦﷮2﷯﷮4﷯ − ﷐﷐𝑥﷮2﷯﷮9﷯ = 1 The above equation of hyperbola is of the form ﷐𝑦2﷮𝑎2﷯ – ﷐𝑥2﷮𝑏2﷯ = 1 ∴ Axis of Hyperbola is y-axis Comparing (1) & (2) a2 = 4 a = 2 & b2 = 9 b = 3 Also, c2 = a2 +b2 Putting value of a2 & b2 c2 = 4 + 9 c2 = 13 c = ﷐﷮𝟏𝟑﷯ Co-ordinates of foci = (0, ±c) = (0, ±﷐﷮13﷯) Foci are (0, ﷐﷮13﷯) & (0, −﷐﷮13﷯) Co-ordinates of vertices = (0, ±a) = (0, ±2) So co-ordinates of vertices are (0, 2) & (0, −2) Eccentricity = e = ﷐𝑐﷮𝑎﷯ = ﷐﷐﷮13﷯﷮2﷯ Latus rectum = ﷐2﷐𝑏﷮2﷯﷮2﷯ = ﷐2 × 9﷮2﷯ = 9

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.